3

This is the proof I read from here. I will quote it fully:

The answer is NO. To see why, consider a line L in the plane P, and two marked points A, B on it. It is desired to construct the midpoint M of the segment AB using the straightedge. Suppose we have found a procedure which works. Now, suppose we have a one-to-one mapping of plane P onto another plane P' which carries lines to lines, but which does not preserve the relation "M is the midpoint of the segment AB", in other words A, M, B are carried to points A', M', B' with A'M' unequal to B'M'. Then, this leads to a contradiction, because the construction of the midpoint in the plane P induces a construction in P' which also would have to lead to the midpoint of A'B'. (This is a profound insight, an "Aha" experience, and worth investing lots of time and energy in thinking it through carefully!!)

I don't understand how the one-to-one mapping induces an equivalent construction of the midpoint of A'B' in the plane P', given that it only preserves lines?.

vladimirm
  • 1,028
  • 1
    The idea is supposed to be that the construction mapped over to $P'$ will construct a point that is not the midpoint of $A'B'$ (as explained in detailed in Brian Scott's answer). This begs the question of what this mapping that preserves lines but not midpoints of line segments might be. According to the link, the mapping "almost" exists. What on earth does that mean? – Rob Arthan Sep 05 '15 at 21:45
  • I'm missing something here ... what if the problem was to find the mid-point using straightedge and compass - this is clearly possible, but as far as I can see the argument still applies, what is the nuance I'm missing about pure straightedge constructions here ? Please enlighten!! – user247608 Sep 05 '15 at 23:05
  • @user247608: presumably this wonderful mapping that preserves lines but not midpoints of line segments doesn't preserve circles. Perhaps that's a clue to understanding its "almost" existence. – Rob Arthan Sep 06 '15 at 00:23
  • Thanks Rob - I'm not feeling very smart right this moment, but transformations which preserve lines presumably must be linear - which would preserve mid-points. Circles would be mapped into 'squashed' circles, ( I'd say ellipses, but what does a shear do?). Is somebody pulling our legs here? This is bugging me, put me out of my anguish! – user247608 Sep 06 '15 at 00:42

2 Answers2

2

If you’re limited to a straightedge, every step of your construction in $P$ must be drawing the line through some specified pair of points. Each of these lines is carried by the mapping to the line of $P'$ through the images of those points. Thus, the whole construction is carried to a construction that yields $M'$ from $A'$ and $B'$ instead of their actual midpoint. Yet it’s the same construction, in the sense that the points used to determine each successive line of the construction are determined by the same rule applied to $A'$ and $B'$, so if it really worked, it would have to construct the actual midpoint of $\overline{A'B'}$.

Brian M. Scott
  • 616,228
  • I don't understand how it's the same construction. If there is a rule that is applied to A and B to draw the next line, how do we know that the transformation of that line from P to P' and applying the same rule on A' and B' will yield the same line in P'? – vladimirm Sep 06 '15 at 00:22
  • @vladimirm: By induction. Any line that you draw during the construction is generated by two points that you've already constructed. Suppose that at some stage the points are $C$ and $D$. By hypothesis they get mapped to some points $C'$ and $D'$ that are correctly related to the images of all earlier lines and points of the construction. The next line drawn in $P$ is the one through $C$ and $D$. It gets mapped to a line in $P'$, and that line contains $C'$ and $D'$ (since they're the images of $C$ and $D$), so it gets mapped to the line generated by $C'$ and $D'$. By induction the whole ... – Brian M. Scott Sep 06 '15 at 06:51
  • ... construction in $P$ gets mapped to the corresponding construction in $P'$. – Brian M. Scott Sep 06 '15 at 06:52
0

It depends on your "initial position". Given A, B, and possibly other points on the line AB (but not M) we cannot obtain other points on AB unless we are given at least 2 points C,D not on AB. Now suppose M lies on the line CD. If we have a projection ,that is, a perspective view P' of P, the recipe "Take the intersection M of CD with AB" transforms to "Take the intersection of A'B' with C'D' " which may or may not give the mid-point of A'B' because projections in general do not preserve ratios of lengths and do not in general preserve mid-points,, but M is still the mid-point of AB . I don't know which book you refer to.I don't intend to read it.

DanielWainfleet
  • 57,985